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Suppose that $f:\mathbb{D}\to \mathbb{C}$ is holomorphic with $$2\left|f'(0)\right|=\sup_{z,\omega \in \mathbb{D}} \left|f(z)-f(\omega)\right|$$ Prove that $f$ is linear.

My attempt

Suppose that $$f(z)=\sum_{n=0}^\infty a_n z^n$$ Then $2\left|f'(0)\right|=2|a_1|$, and \begin{align*} \sup_{z,\omega \in \mathbb{D}} \left|f(z)-f(\omega)\right| &\ge \sup_{|z|=1} \left|f(z)-f(-z)\right| \\ &=2|a_1|\sup_{|z|=1} \left|1+a_3z^2+a_5z^4+\cdots\right| \\ &\ge 2|a_1| \end{align*} where we use the maximum modulus principle.

But I got stuck to show that the equality holds precisely when $a_2=a_3=\cdots=0$. Any hints would be highly appreciated.

Chiquita
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3 Answers3

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Assume wlog $f'(0)=a_1=1$ so $|f(z)-f(w)| \le 2$

Let $f(\mathbb D)=U$ open; consider the convex open hull $K$ of $U$. In other words, we define inductively $U_1=U, U_2$ the union of segments with endpoints in $U_1$, $U_3$ the union of segments with endpoints in $U_2$ etc and $K=\cup U_k$ convex since any two points in $K$ appear in some $U_m$ hence the segment joining them is in $U_{m+1}$ hence in $K$.

It is easy to verify that indeed $U_k$ hence $K$ open (so being convex it is simply connected) and that the diameter $d(U_k) \le 2$ by induction using that in a convex quadrilateral, any interior segment is not larger than the maximum of the sides and diagonals, hence $d(K)\le 2$.

But now if $g$ is the unique Riemann map $g:\mathbb D \to K, g(0)=0, g'(0)>0$, it immediately follows by Schwarz Lemma applied to $g^{-1}(f(z))$ that $g'(0) \ge 1$ and by the diameter property $g'(0) \le 1$, hence $g'(0)=1, f=g$ convex univalent and $U=K$

It is well known and not hard to prove by symmetrization that the area of $K$ is at most $\pi d(K)^2/4=\pi$

But by the usual integral formula since $f(z)=z+\sum_{k\ge 2}{a_kz^k}$ univalent, we get that the area of $K$ is $\int_{\mathbb D}|f'(z)|^2dxdy=\pi(1+\sum_{k\ge 2}{k|a_k|^2})$ hence $a_k=0, k \ge 2$ and we are done!

Conrad
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  • (+1) That's a beautiful proof! :) – r9m Apr 12 '20 at 07:02
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    @r9m - I tried for a while to get an analytic solution and couldn't so then I thought of this as symmetrization methods are powerful in problems about the conformal radius at a point and convex symmetrization works while it is easier to explain than Steiner symmetrization say (same proof works but it is a more difficult result that the conformal radius increases under Steiner symmetrization as in general $U$ is not included in the Steiner symmetrization $V$ with respect the real axis say which is simply connected and preserves diameter) – Conrad Apr 12 '20 at 07:08
  • Wow! I didn't know there were such theorems of symmetrization with conformal maps!! – r9m Apr 12 '20 at 07:25
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    @r9m I agree that something like that should work and it probably is not that hard to make rigorous at least up to the area inequality (here for smooth Jordan curves, the result is doable, while for bounded nice convex Jordan curves it is elementary - grade school geometry - but needs some work); the one issue would be to show that one has $f$ univalent though as otherwise those domains can be ugly and that again probably follows from the maximality of the conformal radius $|f'(0)|$ given the diameter of $f(\mathbb D)$ – Conrad Apr 12 '20 at 15:44
  • nvm the previous comment, it needs more work. I am not sure about it .. – r9m Apr 12 '20 at 16:10
  • Borrowed your symmetrization idea and added a slight variation of the proof :) – r9m Apr 12 '20 at 21:38
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    @r9m - very nice and I think it makes it clearer why the result holds – Conrad Apr 12 '20 at 21:46
  • The OP, as clarified in the comments to the question, uses $\mathbb{D}$ for the closed unit disk. I think that an elementary proof can be found with this stronger hypotesis. –  Apr 13 '20 at 06:33
  • @Caffeine - I am not sure that matters that much since $f$ exists and is conformal ae on the unit circle by the hypothesis and some theory, but if you find one such it would be great to post it as personally I am always interested in cool ideas about this - the fundamental reason the problem holds is that any domain with diameter $2$ cannot have conformal radius $1$ (or higher) at any point, unless it is a disc centered at precisely that unique point - by elementary geometry one can reduce to the convex case and use the diameter/area inequalities, while the complex analysis is just Schwarz – Conrad Apr 13 '20 at 15:12
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Let, $\displaystyle f(z) = \sum\limits_{k=0}^{\infty} a_kz^k$ for $z \in \mathbb{D}$. Then defining $d_r := \operatorname{diam}f(r\mathbb{D})$ we see that $d_r/r$ is a non decreasing function in $r$ (this follows by applying the max-modulus principle to $\displaystyle \frac{f(z) - f(wz)}{z}$ on the disk $z \in r\mathbb{D}$ where, $|w| = 1$). Therefore, letting $r \to 0^+$ we see that $$2|a_1| = 2|f'(0)| = \limsup\limits_{r \to 0^+} \frac{d_r}{r} \le d_1. \tag{1}$$ Then the equality $2|a_1| = d_1$ implies $d_r/r = 2|a_1|$ for all $r \in [0,1)$.

Also, from Schwarz lemma we must have $$|f(z) - f(-z)| \le \frac{d_r}{r}|z|, \, \text{ for } z \in r\mathbb{D} \tag{2}$$ and in particular $2|f'(0)| = 2|a_1| \le d_r/r$ (letting, $|z| \to 0^+$). I.e., the equality $2|a_1| = d_r/r$ corresponds to equality in Schwarz lemma. Hence, we have $$f(z) - f(-z) = \frac{d_r}{r}z = 2a_1z \tag{3}$$ (wlog, we can assume that the unimodular constant is $1$ and in particular $a_1$ is a real number).

Now, consider the function $g(\theta) := |f(e^{i\theta}z) - f(-z)|^2$ where, we have fixed a $z \in \partial r\mathbb{D}$ (i.e., $|z| = r$). Then from $(2)$ we know that $g(\theta)$ is maximized when $\theta = 0$. In particular we must have $g'(0) = 0$.

Now, substituting from relation $(3)$ we note that \begin{align*}g'(\theta) &= \frac{d}{d\theta}\left|f(e^{i\theta}z) - f(z) + 2a_1z\right|^2 \\&= \frac{d}{d\theta} \left[\left(f(e^{i\theta}z) - f(z) + 2a_1z\right) \left(\overline{f(e^{i\theta}z) - f(z) + 2a_1z}\right) \right] \\&= 2 \Re \left[ ie^{i\theta}zf'(e^{i\theta}z)\left(\overline{f(e^{i\theta}z) - f(z) + 2a_1z}\right) \right]. \tag{4}\end{align*}

That is $\displaystyle g'(0) = -2|z|^2a_1\Im \left[f'(z)\right] = 0$ for all $|z| = r$, and hence $\Im \left[f'(z)\right] = 0$ for $|z| = r$ implies $f'(z) \equiv a_1$ in $\mathbb{D}$. That is $f(z) = a_0 + a_1z $ is a linear function. $\square$

An alternative approach: (inspired by Conrad's solution)

Let us denote $\displaystyle N(r) := \frac{1}{\pi r^2} \int_{r \mathbb{D}} |f'(z)|^2\,dx\,dy$ for $r \in [0,1]$.

Now, note that $\lim\limits_{r \to 0^+} N(r) = |f'(0)|^2 > 0$ (since, $f'(0) \neq 0$ otherwise it is trivial), i.e., since $f$ is locally injective near origin by the Area formula we have $$\frac{\text{Area}(f(\mathbb{rD}))}{\pi r^2} = N(r) = \frac{1}{\pi r^2}\int_{\mathbb D} |f'(z)|^2 \,dx\,dy = \sum_{k=1}^{\infty} k|a_k|^2r^{2k-2}$$ for all $r$ small enough. Therefore, $N(r)$ is strictly increasing for small $r > 0$ unless $a_k = 0$ for all $k \ge 2$, i.e., $f$ is linear.

Coming back to the problem if we assume $f$ is not linear then for small $r > 0$, $$|f'(0)|^2 = N(0) < N(r) = \frac{\text{Area}(f(\mathbb{rD}))}{\pi r^2} \le \frac{\pi d_r^2}{4\pi r^2} = |f'(0)|^2 \tag{5}$$ where, the second inequality in $(5)$ is due to the isodiametric inequality followed by the equality established in eqn $(1)$. Contradiction!

Hence, $f$ must be linear. $\square$

r9m
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  • very nice solution and in the spirit of the problem (+1) - I added a geometric solution (as I couldn't think of the analytic one above) – Conrad Apr 12 '20 at 06:46
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One can also prove the result withouth resorting to geometric methods and instead using $2$ complex variables:

Define $$F(z,w)=\begin{cases}\frac{f(z)-f(w)}{z-w} &z\neq w \\ f'(0)&\text{otherwise}\end{cases}$$ It is easy to see that $F(z,w)\in A(\mathbb{D}^2)$ (the algebra of holomorphic functions on the polydisc continuous up to the boundary). Now, for functions in $A(\mathbb{D}^2)$ a particularly strong version of the maximum principle holds: the maximum of $|F(z,w)|$ is obtained on $\mathbb T^2$ (i.e. $\{(z,w)\in\mathbb D^2: |z|=|w|=1\}$). This version of the maximum principle implies that $2|f'(0)|\le \max_{z,w\in \partial \mathbb{D}}|f(z)-f(w)|$. If equality holds the maximum principle implies that $F$ is constant proving the claim.

The claimed version of the maximum principle is proved in "function theory in the polydiscs" by W. Rudin. There is also an elementary proof of the result: by the Cauchy integral formula it follows that, given $g\in A(\mathbb{D}^2)$ and $(z_0,w_0):|z_0|,|w_0|<1$ we have $$|g(z_0,w_0)|\le \frac{1}{d(z_0,\partial \mathbb{D})d(w_0,\partial \mathbb{D})}\max_{(z,w)\in \mathbb T^2} |g(z,w)|$$ Since the formula holds for any $g\in A$, it holds for $g^n$ too, so $$|g(z_0,w_0)|\le K(z_0,w_0)^{\frac1n}\max_{(z,w)\in \mathbb T^2} |g(z,w)|$$ taking $n\to \infty$ we get the claim.